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I've manged to prove that all groups of order $< 60$ are solvable, using Burnside's theorem. However, I found an alternate proof here Question about solvable groups It states that:

"Note that $A_5$ is the smallest non-abelian simple group and its order is $60$. Therefore in any subnormal series of any group of order less than $60$, $A_5$ is not a composition factor. Hence all group of order less than $60$ are all solvable."

Could anyone explain why the fact that in any subnormal series of any group of order $< 60$, $A_5$ is not a composition factor implies that all groups of order less that $60$ are solvable? It's not clear to me why that follows immediately.

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  • $\begingroup$ Well if $A_5$ is the smallest nonabelian simple group, then all composition factors of order less than $60$ must be abelian simple groups - i.e. they must be cyclic of prime order, so the group is solvable. But this is not really an alternative proof, because proving that $A_5$ is the smallest nonabelian simple group is essentially equivalent to proving that all groups oforder less than $60$ are solvable. $\endgroup$ – Derek Holt Oct 22 '14 at 6:26
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Let $N$ be a maximal normal subgroup. Then $G/N$ is simple and abelian (because it is smaller than $A_5$), hence cyclic. $N$ is strictly smaller than $G$, hence we may assume by induction that itis solvable, hence $G$ is solvable.

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